In the answer of a question Difference between an isometric operator and a unitary operator on a Hilbert space states:
On infinite dimensional Hilbert spaces (unlike in finite dimensional cases), there are always nonunitary isometries.
So, I'm thinking that something must be wrong with the finite dimensional example that occurs to me? I'd appreciate feedback on this…
Consider a canonical injection of $T: R → R^2$ defined by $T(x) = (x, 0)$ with the usual Euclidean metrics. $T$ is linear, isometric, and not surjective.
Define $T^*: R^2 → R$ by $T^*(x, y) = x$. Then
- $T^*$ is the (unique) adjoint of $T$: $〈 T(x), (x`, y`)〉 = x.x` = 〈 x, T^* (x`, y`)〉$
- $T^*$ is the left inverse of $T$: $T^*( T(x)) = T^*(x, 0) = x,$ so $T^*( T(.)) = I.$
But for $y` ≠ 0$, $T( T^*(x`, y`)) = T(x`) = (x`, 0) ≠ (x`, y`)$, so $T( T^*(.)) \not = I,$ and therefore T is not unitary.
On reflection, I think the post cited refers specifically to operators on a single space rather than general linear transformations between different spaces. Can anyone confirm this is the cases and that my example is fine for an isometric transformation between different spaces ?
Best Answer
Yes, I meant from a space to itself. An injective linear operator on a finite dimensional space (that is, to and from the same space) must be surjective. Your example is a good one for a non-surjective isometry between different spaces.